Let the probability of success on Bernoulli trial be 026 a I

Let the probability of success on Bernoulli trial be 0.26

a) In four Bernoulli trials, what is the probability that there will be 3 failure ( Answer 4 Decimal Placess)

b) In four Bernoulli trials, what is the probability that there will be more than the expected number of failures (Answer 4 Decimal places)

Solution

A)

Three failures = 1 success.

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    4      
p = the probability of a success =    0.26      
x = the number of successes =    1      
          
Thus, the probability is          
          
P (    1   ) =    0.42143296 [ANSWER]

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B)

The mean number of successes is

mean success = n p = 4*0.26 = 1.04

Thus,

mean failures = n(1-p) = 2.96

Thus, it exceeds the mean number of failures if there are 3 or 4 failures, or, at most 1 success.

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    4      
p = the probability of a success =    0.26      
x = the maximum number of successes =    1      
          
Then the cumulative probability is          
          
P(at most   1   ) =    0.72129872 [ANSWER]